Lemma 2.1 [1]}, [2]}
For any function \(x: \mathbb {N}_{a+1-n} \rightarrow \mathbb {R}\) , \(n \in \mathbb {Z}_{+}\) , \(a \in \mathbb {R}\) , one has
\({\textstyle { }_{a}^{\rm R} \nabla _{k}^\alpha {}_ a^{ \rm G} \nabla _{k}^{-\alpha } x(k)={ }_{a}^{\rm C} \nabla _{k}^{\alpha }{}_a^{ \rm G} \nabla _{k}^{-\alpha } x(k)=x(k),}\) 
\({\textstyle { }_{a}^{\rm G} \nabla _{k}^{-\alpha } {}_a^{\rm R} \nabla _{k}^{\alpha } x(k)=x(k)-\sum _{i=0}^{n-1} \frac{(k-a) \overline{^{\alpha -i-1}}}{\Gamma (\alpha -i)}[{ }_{a}^{\rm R} \nabla _{k}^{\alpha -i-1} x(k)]_{k=a},}\) 
\({\textstyle { }_{a}^{\rm G} \nabla _{k}^{-\alpha }{ }_{a}^{\rm C} \nabla _{k}^{\alpha } x(k)=x(k)-\sum _{i=0}^{n-1} \frac{(k-a) \overline{^i}}{i !}[\nabla ^{i} x(k)]_{k=a} , }\) 
